Zapisz w postaci iloczynowej
y=xkwadrat+2x-1
y=xkwadrat-3x-2​


Odpowiedź :

Funkcja kwadratowa jej postacie :

f(x) = ax² + bx + c     postać ogólna

f(x) = a× ( x - p )² + q   postać kanonicza  gdzie W= ( p , q ) są to współrzędne wierzchołka

f(x) = a × ( x - x₁ ) × ( x  - x ₂ )   postać iloczynowa gdy  Δ > 0

f(x) = a × ( x - x₀ )²   postać iloczynowa gdy Δ = 0

[tex]zad.1\\\\y=x^{2} +2x-1\\a=1,~~b=2,~~c=-1\\\Delta = b^{2} -4ac\\\\\Delta=2^{2} -4\cdot 1\cdot (-1)=4+4=8\\\sqrt{\Delta} =\sqrt{8} =2\sqrt{2} \\\\x_{1} =\dfrac{-b-\sqrt{\Delta} }{2a} ~~\lor ~~x_{2} =\dfrac{-b+\sqrt{\Delta} }{2a}\\\\\\x_{1} =\dfrac{-2-2\sqrt{2} }{2}=-1-\sqrt{2} ~~\lor~~x_{2} =\dfrac{-2+2\sqrt{2} }{2}=-1+\sqrt{2}\\\\y=1\cdot (x-(-1-\sqrt{2} ))\cdot (x-(-1+\sqrt{2} ))=(x+1+\sqrt{2} )\cdot (x+1-\sqrt{2} )\\\\[/tex]

[tex]zad.2\\\\y=x^{2} -3x-2\\a=1,~~b=-3,~~c=-2\\\Delta = b^{2} -4ac\\\\\Delta=(-3)^{2} -4\cdot 1\cdot (-2)=9+8=17\\\sqrt{\Delta} =\sqrt{17} \\\\x_{1} =\dfrac{-b-\sqrt{\Delta} }{2a} ~~\lor ~~x_{2} =\dfrac{-b+\sqrt{\Delta} }{2a}\\\\\\x_{1} =\dfrac{3-\sqrt{17} }{2} ~~\lor~~x_{2} =\dfrac{3+\sqrt{17} }{2}\\\\y=1\cdot (x- \dfrac{3-\sqrt{17} }{2})\cdot (x- \dfrac{3+\sqrt{17} }{2})\\\\y= (x- \dfrac{3-\sqrt{17} }{2})\cdot (x- \dfrac{3+\sqrt{17} }{2})[/tex]